$R^1$ and $H^1$ for a scheme over a field

Question

Let $S=Spec(k)$ for a field $k$ and $\;X\overset{f}\longrightarrow S$ a scheme over $S$. Is it correct to say that $$ R^1 f_* \mathcal{O}_X \cong H^1(X, \mathcal{O}_X) \quad ? $$

Here $R^1f_*$ is the right derived functor of the direct image functor $\;f_*$ and $H^1(X, \mathcal{O}_X)$ denotes the first Chech cohomology group of the structure sheaf of $X$.

Intuitively I think the above holds... maybe under some hypothesis on $X$?

Anyway I'm a bit confused, so any correction or suggestion is very welcome!

Answer 1

Here's the answer to my (rather trivial) question, thanks Asal for the advice.

Let $F$ be a sheaf on $\;X \overset{f}\longrightarrow S$, with $S=Spec(k)$. By definition the direct image functor associates to $F$ the sheafification of the pre-sheaf $$ U \mapsto F(f^{-1}(U)). $$ Now, since $S$ is just a point, the only open $U$ is $S$ itself, and its preimage $f^{-1}(S)$ is the whole $X$. Hence we have $f_*F$ is a sheaf defined on a one point space with stalk $F(X)$, therefore $f_*$ can be identified with the global sections functor $\Gamma$.

Hence the right derived functors of $f_*$ and $\Gamma$ are the same, and we get $$ R^1f_* (\square) = R^1\Gamma(\square) = H^1(X,\square). $$

Remark: This holds if the right derived functor of $\Gamma$ agrees with the first Cech cohomology functor. This is not the case in general.